Question

An automobile travelling at $50\, kmh^{-1}$, can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 90 $kmh^{-1}$, all other conditions remaining same and assuming no skidding, the minimum stopping distance in metre is

• A. 72
• B. 92.5
• C. 102.6
• D. 129.6

Answer 1

Answer: (D)

129.6

Answer 2

By equation of motion $v^2 - u^2$ = 2as where u is initial velocity, a is acceleration and s is displacement. Given $u = 50 kmh^{-1},v = 0, = 40 m$ hence $a=\frac{u^2}{2s}=\frac{\left(50 \times \frac{5}{18}\right)^2}{2 \times 40}$ when u' = $90 kmh^{-1}$ so,s=$\frac{u'^2}{2a}$ $\Rightarrow$ $\frac{\left(90 \times \frac{5}{18}\right)^2 \times 2 \times 40}{2 \times \left(50 \times \frac{5}{18}\right)^2}$ or $s=129.6 \,m$